A non-reflexive Banach space with all contractions mean ergodic

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1 A o-reflexive Baach space with all cotractios mea ergodic Vladimir P. Fof, Michael Li Be-Gurio Uiversity Przemyslaw Wojtaszczyk Uiversity of Warsaw May 4, 2009 Dedicated to the memory of Aryeh Dvoretzky Abstract We costruct o ay quasi-reflexive of order separable real Baach space a equivalet orm, such that all cotractios o the space ad all cotractios o its dual are mea ergodic, thus aswerig egatively a questio of Louis Suchesto. Itroductio A liear operator T o a (real or complex) Baach space X is called mea ergodic if the limit E(T )x := lim T k x exists x X. k= Mea ergodicity was show by Vo-Neuma (93) for uitary operators o Hilbert spaces, by Riesz (937) for cotractios o Hilbert spaces, ad (idepedetly) by Lorch (939), Kakutai (938) ad Yosida (938) for power-bouded operators o reflexive spaces. We refer the reader to [9] for proofs, discussio ad refereces. A atural questio is whether reflexivity is ecessary. I [6] the authors proved that if every power-bouded operator o a Baach space with a basis is mea ergodic, the the space must be reflexive; it is still ot kow if the same holds without the existece of a basis. For additioal refereces related to this questio see [6]. Suchesto [8] posed the followig questio: If every cotractio i a Baach space is mea ergodic, must the space be reflexive? I this paper we costruct a example which gives a egative aswer to Suchesto s questio.

2 2 Operators o quasi-reflexive spaces of order Defiitio A Baach space X is quasi-reflexive of order if dim X /X =, where we always cosider X X via the caoical isometric embeddig. The first example of such a space (over R, separable with a basis), ad oe of the importat oes, is the James space [7] (for more of its properties see [4]). Throughout this ote we cosider Baach spaces over R. For the sake of completeess we iclude the followig result of [3, 2.2]. Propositio. Let X be quasi-reflexive of order. The there exists a liear, multiplicative fuctioal q : L(X) R of orm such that ker q is exactly the space of weakly compact operators from X ito X. Proof. For T : X X let us cosider T : X X. Sice T (X) X we see that it iduces a operator T : X /X X /X. It is easy to check that T T = T. Sice X /X is oe-dimesioal, T is a multiplicatio by a umber which we deote by q(t ). It is clear that the map q is the desired orm liear multiplicative fuctioal. The kerel of q cosists of those operators T : X X such that T (X ) X, but this is exactly the set of all weakly compact operators. Now for T : X X we write T = q(t )I+(T q(t )I). Sice q (T q(t )I) = 0, by the propositio W := T q(t )I is weakly compact. If T : X X is power-bouded we get c T q(t ) = q(t ) for =, 2,... so q(t ). Let us summarize this discussio with Corollary. Every liear operator T o X quasi-reflexive of order has a uique decompositio as T = λi + W where λ R ad W is weakly compact. If T is power-bouded, the λ. Propositio 2. Let X be quasi-reflexive of order. Suppose T = λi + W : X X is power-bouded, with λ. The both T ad T are mea-ergodic. Proof. For the space of fixed poits of T, deoted by Fix (T ), we have that Fix (T ) = {x X : λx + W (x) = x} = {x X : W (x) = ( λ)x} is a eigespace correspodig to a o-zero eigevalue of a weakly compact operator, so it is a reflexive space. Aalogous argumets show that Fix (T ) = {x X : W (x ) = ( λ)x }. Sice W is weakly compact, we have W (X ) X, which implies Fix (T ) = Fix (T ). It is kow that Fix (T ) always separates Fix (T ) (e.g. use [9, Theorem 2..3, p. 73]), so i our case we get that Fix (T ) separates Fix (T ). By Sie s criterio [7], [9, p. 74], we obtai that T is mea-ergodic. But aalogously Fix (T ) always separates Fix (T ), so i our case Fix (T ) separates Fix (T ), ad by Sie s criterio T is mea-ergodic. 2

3 Remark. It was show i [6, Theorem 5] that for T power-bouded o a quasi-reflexive space of order we always have T or T mea ergodic, ad if the space has a basis there exists T power-bouded which is ot mea ergodic; by the previous propositio this T is of the form I + W with W weakly compact. 3 Reormig spaces with separable secod dual I this sectio X is a Baach space over R. We use the stadard otatios S X = S (X, ) = {x X : x = } ad B X = B (X, ) = {x X : x } for the uit sphere ad the closed uit ball of X, respectively. Propositio 3. Let X be a o-reflexive Baach space with X separable. The there exist a equivalet orm o X, a fuctioal f 0 S (X, )\X, ad a fuctioal F 0 S (X, ) X such that (i) F 0 (f 0 ) = ad F 0 (g) < for ay g B (X, ), g f 0. (ii) If H S (X, ) ad H(f 0 ) =, the H X. Moreover, if there exists a Baach space Y such that X = Y (isometrically), the the orm ca be take as a dual orm. Proof. We prove first the case that X = Y ; we will the idicate how to modify the proof for the o-dual case. As usual we assume that Y Y = X ad X = Y Y = X, uder the caoical isometric embeddigs. Claim. For ay y S Y Y, we have d(y, Y ) /2. Proof of Claim. For y Y we have y y sup y (y y )(y) = y. Hece y y max{ y y, y } 2 y = 2, which proves the claim. We costruct a ew orm i 2 steps. Pick a fuctioal f S Y. Clearly, f S X \ X, ad by Claim d(f, X) /2. By the Hah-Baach theorem there is F S X X such that F (f) = d(f, X) /2. Next take a sequece {x } S X such that w lim x = f (i the w -topology of X ). Put W = cl co{b X {±3x } =}, W = w clw. Sice f Y it follows that w lim x = 0 i w -topology defied o X by its predual Y. Sice B X is w -compact (i the w -topology defied o X by the predual Y ), it easily follows that W is w -compact (ideed, sice w lim x = 0, it follows that A = cl co{±3x } is w -closed, ad hece W = co{a B X } is w -compact, by [3, Lemma V.2.5]). By Milma s theorem [2] (see [6, Prop..5]), extw extb X {±3x } = {±3f}. () Sice X is separable, W is a weak-* compact covex set with extw separable, so by [8],[5] (see also [6, p. 26]), we have that W = cl co{extw } the orm closed covex hull (this ca be deduced also from the Bessaga-Pe lczyński theorem [2]). It follows that supf (W ) = supf (extw ). 3

4 Sice F = ad F X, by () we have supf (W ) = F (3f) >, ad, moreover, F (g) < F (3f) for ay g W, g 3f. Let {u i } i= B X be a sequece dese i B X. Put f 0 = 3f, t = 3x, F 0 = F (3f) F, K = cl co{±2 i u i } i=, T = {λt + x : x K, λ 2 + x 2 }, =, 2,..., T 0 = {λf 0 + x : x K, λ 2 + x 2 }, Sice K is symmetric, each T is symmetric, ad we defie V = cl co{w =T }, V = w clv. Claim 2. V is w -closed i the w -topology o X defied by the predual Y. Proof of Claim. Put A = w cl co = T. A easy cosideratio shows that w cl = T = =T K (recall that w lim t = 0). Sice A is a weak-* compact covex set (i X = Y ), by Milma s theorem exta w cl = T = =T K, so by [8],[5] A = cl co{ =T K}. Sice K B X W, it follows that A V. Fially, sice also W is w -compact ad covex, we obtai (usig [3, Lemma V.2.5]) V w cl co{w A} = co{w A} V, hece V = w cl co{w A} is w -closed, which fiishes the proof of Claim 2. Sice V is a bouded closed covex symmetric subset of X cotaiig B X, it is easy to show that x := if{t > 0 : t x V } defies a equivalet orm o X with B (X, ) = V. From Claim 2 we get that is a dual orm (see, e.g., [9]). Next, by Milma s theorem (i the weak-* topology of X ) extv W =0T, ad sice V is weak-* compact covex, it is the orm-closed covex hull of extv ([8],[5]), ad by the choice of F 0 (i particular, F 0 X ), we have supf 0 (V ) = sup F 0 (extv ) = sup F 0 (W ) = F 0 (f 0 ) =. Moreover, f 0 is the oly poit i W =0T where F 0 attais the value. Ideed, for W it was already metioed above, while for the set =0T it easily follows from F 0 X. Fially assume that H X satisfies H(f 0 ) = = max H(V ); we prove that H X. Fix y K ad put D = {af 0 + by : a 2 + b 2 }. We show that D T 0 : For af 0 + by D we have a 2 + b 2. Sice K B X is absolutely covex, y K ad b imply that x = by K, ad y yields a 2 + x 2 a 2 + b 2. Therefore af 0 + by T 0 V. Thus we have = H(f 0 ) = max H(D). 4

5 Let H(y) = γ. We the have ( + γ 2 ) (f 0 + γy) D ad H ( + γ 2 f 0 + γ + γ 2 y) = + γ 2 = + γ 2 + γ 2 + γ 2 so we must have H(y) = 0. Sice y K is arbitrary ad cl spak = X, it follows that H X. Whe X is ot a dual space, we skip Claim ad start directly by takig f S X \ X with d(f, X) > /3, ad the procced with the same proof, igorig (the ow uecessary) Claim 2. Remark. Clearly, the coditio X is separable may be weakeed. For istace, the same proof works, usig deeper results, if X is separable ad does ot cotai l (use the remark at the ed of [5], istead of [8],[5] or [2], for expressig W ad V as the orm-closed covex hull of their respective extreme poits). Corollary 2. Let X be a separable quasi-reflexive of order Baach space. The there exist a equivalet orm o X, a fuctioal f 0 S (X, )\X, ad a fuctioal F 0 S (X, ) X, such that (i) F 0 (f 0 ) = ad F 0 (g) < for ay g B (X, ), g f 0. (ii) F 0 is the oly fuctioal i S (X, ) with F 0 (f 0 ) =. Moreover, if there exists a Baach space Y such that X = Y (isometrically), the the orm ca be take as a dual orm. Proof. For part (ii), ote that X = X [f 0 ] by quasi-reflexivity of order. 4 Mea ergodicity of cotractios I this sectio we costruct o-reflexive separable Baach spaces such that every cotractio is mea ergodic (abbreviated ME i the sequel). We start with a geeral lemma o mea ergodicity. Lemma. Let R be a mea ergodic power-bouded operator o a Baach space Y, with ergodic projectio Ey = lim k= Rk y. The E (Y ) = Fix (R ). Proof. Sice RE = ER = E = E 2, we have R E = E, which yields E (Y ) Fix (R ). For the coverse, let y Fix (R ). The for y Y we have E y (y) = y (Ey) = y (lim R k y) = y (y). k= Throughout this sectio, we assume that X is a separable Baach space quasi-reflexive of order, edowed with the orm give by Corollary 2, ad we use the otatios of Corollary 2. We will show that every cotractio o X is mea ergodic. 5

6 Lemma 2. Let Q be a orm projectio from (X, ) oto the oedimesioal subspace spaed by f 0. The Qx = F 0 (x )f 0. Proof. Clearly Qx = α(x )f 0, with α liear (by liearity of Q), ad α = sice α(x ) = Qx x with equality o Q(X ) {0}. Now α(f 0 )f 0 = Qf 0 = Q 2 f 0 = α 2 (f 0 )f 0 so α(f 0 ) = sice Q 0. By Corollary 2(ii), α = F 0. Lemma 3. Let T = I+W be a cotractio i (X, ) with W weakly compact. The f 0 Fix (T ). Proof. Sice W is weakly compact, it follows that W f 0 F 0 (W f 0 ) = 0. Hece X, ad hece T f 0 = f 0 + W f 0 F 0 (f 0 ) + F 0 (W f 0 ) = + F 0 (W f 0 ) =. Therefore f 0 + W f 0 =, ad we coclude that F 0 attais its orm o f 0 + W f 0. Corollary 2(i) yields that W f 0 = 0, i.e. f 0 Fix (T ). Theorem. Every separable Baach space which is quasi-reflexive of order has a equivalet orm i which ay cotractio is mea ergodic. Proof. Edow X with the orm obtaied i Corollary 2. Let T : X X be a cotractio. By Propositio 2 we have to prove oly the case T = I + W where W is weakly compact (to which Lemma 3 is applicable). By [6], T or T (or both) is ME, so without loss of geerality we may assume that T is ME. Besides, we ca assume that Fix (T ) {0} (otherwise X = (I T )X so T is ME ad we are doe). Defie P x = lim T k x, x X. i= Clearly, P is a projectio oto Fix (T ), ad sice Fix (T ) {0} ad T, it follows that P = (i.e. P is ot 0). Sice T is ME we have the followig ergodic decompositio X = P (X ) (I T )X = F (T ) W X. By Lemma 3 f 0 Fix (T ), so Fix (T ) = Fix (T ) [f 0 ]. Defie Qx = F 0 (P x )f 0. The Qf 0 = f 0 ad Q =, Q 2 = Q. Lemma 2 yields that Qx = F 0 (x )f 0. Hece KerQ = KerF 0 = X (we use here that X is quasi-reflexive of order ). Therefore KerP X. By Lemma Fix (T ) = P (X ), ad X = P X KerP = Fix (T ) KerP = [f 0 ] Fix (T ) kerp, ad hece X = Fix (T ) KerP (2) 6

7 Mea ergodicity of T implies P x = w lim T k x. If P x = 0 the x = x X, ad we have k= T k x 0 weakly, hece i orm (e.g. [9, p. 72]). The decompositio (2) yields that T is ME. Theorem 2. Let X be quasi-reflexive of order, with the orm defied by Corollary 2, ad let T be a cotractio o (X, ); the T is mea ergodic. Proof. By Propositio 2 it remais to prove the theorem oly for T = I + W with W weakly compact, which we ow assume. From Theorem we kow that T is mea ergodic, ad deote Ex := lim k= T k x. The E is a projectio oto Fix (T ), ad E 2 = E = ET = T E. Hece by Lemma E is a projectio of X oto Fix (T ), ad E projects X ito Fix (T ), so we have E f 0 Fix (T ). By Lemma 3 f 0 Fix (T ), ad the decompositio X = X [f 0 ] yields that T is mea ergodic. The ergodic decompositio of X by mea ergodicity of T yields k= X = X [f 0 ] = Fix (T ) (I T )X [f 0 ] (3) so Fix (T ) = Fix (T ) [f 0 ]. Hece E f 0 = λf 0 + y 0 with y 0 Fix (T ). Sice E X = E, we have E y 0 = y 0 X. The fuctioal F 0 is i X, so we have λ = F 0 (λf 0 +y 0 ) = F 0 (E f 0 ) = F 0 (E E f 0 ) = F 0 (λ 2 f 0 +λy 0 +E y 0 ) = λ 2. Case (i): λ =. I this case F 0 (E f 0 ) =, ad sice f 0 = ad E, Corollary 2(i) yields E f 0 = f 0. We have observed that T is mea ergodic with the decompositio (3). Sice E X = E ad E f 0 = f 0, we obtai that lim k= T k x = E x. Hece for x X we have that for every x X x ( T k x ) = ( T k x )(x ) E x (x ) = x (E x ). k= k= Thus k= T k x coverges weakly to E x, ad therefore i orm [9, p. 72]. Hece T is mea ergodic. Case (ii): λ = 0. I this case we have E f 0 Fix (T ), so E (Fix (T )) Fix (T ) (ad equality holds). Let y Fix (T ), ad put y = E y Fix (T ). Sice (for ay cotractio o a Baach space) Fix (T ) separates Fix (T ), there exists y Fix (T ) such that y (y) 0. By Lemma E y = y, which yields y (y ) = y (E y ) = E y (y ) = y (y) 0. Hece Fix (T ) separates Fix (T ), so by Sie s criterio T is mea ergodic. Lemma 4. Let X be a Baach space such that all cotractios are mea ergodic. The every strogly cotiuous semi-group of cotractios {T t } t 0 o X is mea ergodic, i.e. lim t t t 0 T s x ds exists x X. 7

8 Proof. For the sake of completeess we idicate the stadard proof. For x X put y = 0 T sx ds. By the semi-group property T k = T k ad we obtai t t 0 T s x ds = [t] t [t] T k y + [t] t k=0 by mea ergodicity of the cotractio T. t [t] T s x ds E(T )y Remark. Mugolo [4] gave a semi-group aalogue of the result of [6]. We ca ow reiforce our egative aswer to Suchesto s questio. Theorem 3. Every separable Baach space Z which is quasi-reflexive of order has a equivalet orm such that all cotractios o Z ad all cotractios o Z (i the iduced dual orm) are mea ergodic. Moreover, every strogly cotiuous semi-group of cotractios o Z or o Z (i the above orms) is mea ergodic. Proof. Let Z be a separable Baach space quasi-reflexive of order ad let X = Z. The also X is quasi-reflexive of order (e.g. [4, p. 0]), ad by the costructio of the orm o X obtaied i Corollary 2, there is a equivalet orm o Z, deoted by, for which (X, ) = (Z, ). Let R be a cotractio o (Z, ). The T = R is a cotractio o (X, ), ad by Theorem 2 T = R is mea ergodic o X = Z. But R Z = R, so the mea ergodicity of R yields mea ergodicity of R. If T is a cotractio o (Z, ) = (X, ), the by Theorem T is mea ergodic o X. The mea ergodicity of strogly cotiuous cotractio semi-groups follows from the above ad the previous lemma. Browder [, Lemma 5] proved that for T power-bouded o a reflexive Baach space Y we have x (I T )Y if ad oly if sup T k x <. (4) Li ad Sie [] showed that (4) holds also for Y = L ad T ay cotractio (eve ot mea ergodic), ad gave a example of a mea ergodic cotractio T o a subspace of L for which (4) fails. Thus (4) ad mea ergodicity are icomparable. Whe (4) holds we say that T satisfies Browder s coditio. Lemma 5. Let T be a power-bouded operator o a Baach space Y. If T is mea ergodic (o Y ), the (4) holds. Proof. Let sup k=0 T k x <. Browder s result was exteded i [0] (see also []) to show that (4) holds whe Y is a dual space ad T is a dual operator. We apply this to T ad obtai that x (I T )Y. The assumptio that T is mea ergodic yields, by [, Theorem ], that a solutio y Y of the equatio (I T )y = x is give by y = lim N N N = k=0 T k x = lim N which shows that y Y ad x (I T )Y. N k=0 N T k x, = k=0 8

9 Theorem 4. Let Z be a separable Baach space which is quasi-reflexive of order, edowed with the equivalet orm defied i Theorem 3. The every cotractio o Z ad every cotractio o X = Z satisfies Browder s coditio. Proof. Let T be a cotractio o Z. The T is a cotractio o X, ad by Theorem 2 T is mea ergodic o X = Z. Hece T satisfies Browder s coditio by the previous lemma. Now let T be a cotractio o Z = (X, ) ad let x X satisfy sup k=0 T k x < ; the there exists y X with (I T )y = x. Recall (Corollary ) that T = λi + W with W weakly compact (ad λ ). Case (i): λ. Sice W y = z X, we obtai x = (I T )y = ( λ)y W y = ( λ)y z, which yields y = ( λ) (x + z) X, so x (I T )X. Case (ii): λ =. Let y = y + αf 0, with y X. By Lemma 3 T f 0 = f 0, so x = (I T )(y + αf 0 ) = (I T )y. 5 Problems It was proved i [6] that if Y is a Baach space such that every power-bouded operator defied o ay closed subspace of Y is mea ergodic (o that subspace), the Y is reflexive (o basis assumed). The questio (related to aother questio i [8]) is this: If Y is a Baach space such that every cotractio defied o ay closed subspace is mea ergodic, is Y reflexive? If the aswer to the above is egative, what if both Y ad Y have the above property? Ackowledgemets. The research was started durig a visit of the third author to Be-Gurio Uiversity, supported by the Ceter for Advaced Studies i Mathematics of Be-Gurio Uiversity. The research was completed durig a visit of the secod author to the Istitute of Mathematics of the Polish Academy of Scieces (IMPAN) i Warsaw, supported by the EU project TODEQ. The above authors are grateful for the support ad hospitality of their hosts. The third author wishes to thak also the Foudatio for Polish Sciece for its support. Refereces [] F. Browder, O the iteratio of trasformatios i ocompact miimal dyamical systems, Proc. Amer. Math. Soc. 9 (958), [2] C. Bessaga ad A. Pe lczyński, O extreme poits i separable cojugate spaces, Israel J. Math. 4 (966), [3] N. Duford ad J. Schwartz, Liear operators, Part I, Itersciece, New York,

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